Differentiating Logarithmic Functions with Base e

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Derivatives #8: Logarithmic Differentiation

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1 of 2 videos by Darshan Senthil

Key Questions

  • # 1/x # is the derivative of ln(x).


    Let us derive the above derivative by Implicit Differentiation.

    By rewriting in terms of an exponential function,

    #y=lnx Rightarrow e^y =x#

    by implicitly differentiating with respect to #x#,

    #Rightarrow e^ycdot y'=1#

    by dividing by #e^y#,

    #Rightarrow y'=1/e^y=1/x#.

    Hence, #(lnx)'=1/x#.


    I hope that this was helpful.

  • By Quotient Rule,

    #y'={1/x cdot x-lnx cdot 1}/{x^2}={1-lnx}/{x^2}#

    This problem can also be solved by the Product Rule

    #y'=f'(x)g(x)+f(x)g(x)#

    The original function can also be rewritten using negative exponents.

    #f(x)=ln(x)/x=ln(x)*x^-1#

    #f'(x)=1/x*x^-1+ln(x)*-1x^-2#

    #f'(x)=1/x*1/x+ln(x)*-1/x^2#

    #f'(x)=1/x^2-ln(x)/x^2#

    #f'(x)=(1-ln(x))/x^2#

  • The answer would be #f'(x) = 1/g(x)*g'(x)# or it can be written as #f'(x)=(g'(x))/g(x)#.

    To solve this derivative you will need to follow the chain rule which states:

    #F(x)=f(g(x))# then #F'(x)=f'(g(x))*g'(x)#

    Or without the equation, it the derivative of the outside(without changing the inside), times the derivative of the outside.

    The derivative of #h(x) = ln(x)# is #h'(x) = 1/x#.

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Differentiating Logarithmic Functions